\newproblem{lay:1_1_12}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.1.12}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Clara Susana Rey Abad, Oct. 29, 2013} \\}{}

  % Problem statement
	
	Solve the equation system:
	\begin{center}
	 $\begin{array}{rcr}
	  x_1-5x_2+4x_3&=&-3\\
		2x_1-7x_2+3x_3&=&-2\\
		-2x_1+x_2+7x_3&=&-1\\
	  \end{array}$
	\end{center}
	
}{
   % Solution
	Let us construct the augmented system matrix
	\begin{center}
		$\left(\begin{array}{rrr|r}
		   1 & -5& 4 & -3 \\
			 2 & -7& 3 & -2 \\
			-2 & 1 & 7 & -1
		\end{array}\right)$
	\end{center}
	
	Now, we apply row operations to solve it
	\begin{center}
		\begin{tabular}{cc}
			 $\mathbf{r}_2\leftarrow \mathbf{r}_2+\mathbf{r}_3$ &
			 $\left(\begin{array}{rrr|r}
				 1 & -5 & 4  & -3\\
				 0 & -6 & 10 & -3 \\
				-2 &  1 & 7  & -1
			 \end{array}\right)$ \\
			 $\mathbf{r}_3\leftarrow \mathbf{r}_3-2\mathbf{r}_1$ &
			 $\left(\begin{array}{rrr|r}
				 1 & -5  &  4 & -3 \\
				 0 & -6  & 10 & -3 \\
				 0 & -9  & 15 & -7
			 \end{array}\right)$ \\
			
			$\mathbf{r}_2\leftarrow \mathbf {r}_2\div 3$ &
			$\left(\begin{array}{rrr|r}
			   1 & -5 &   4  & -3 \\
				 0 & -2 & 10/3 & -1 \\
				 0 & -9 &  15  & -7
				\end{array}\right)$\\
				
			$\mathbf{r}_3\leftarrow \mathbf {r}_3\div 3$ &
			$\left(\begin{array}{rrr|r}
			   1 & -5 &   4  & -3 \\
				 0 & -2 & 10/3 & -1 \\
				 0 & -3 &   5  & -7/3
				\end{array}\right)$ \\
				
			$\mathbf{r}_3\leftarrow \mathbf {r}_3\cdot 2 -\mathbf {r}_2\cdot 3$ &
			$\left(\begin{array}{rrr|r}
			   1 & -5 &  4   & -3 \\
				 0 & -2 & 10/3 & -1 \\
				 0 &  0 &  0   & -5/3 
				\end{array}\right)$ \\
		\end{tabular}
	\end{center}
	
	Last row represents the equation $0=-5/3$ which is non-sense and, therefore, there is no solution of the system. The equation system is incompatible.
}
\useproblem{lay:1_1_12}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
